we can assume without loss of generality that x and y are of the form x pabcq and y pbcdq,

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1 1. Symmetric and alternating groups. SOLUTIONS Math A4900 Homework 9 11/1/2017 (a) In class we showed that S n is generated by T tpi jq 1 ď i ă j ď nu, the set of transpositions in S n. Show by induction on j i, that for i ă j, pi jq pi i ` 1qpi ` 1 i ` 2q pj 2 j 1qpj 1 jqpj 2 j 1q pi ` 1 i ` 2qpi i ` 1q, and conclude S n is generated by T 1 tpi i`1q 1 ď i ă nu, the set of adjacent transpositions. Proof. We show the claim by induction on j i. When j i 1, the claim is trivially true. Now assume that Then pi ` 1 jq pi ` 1 i ` 2q pj 2 j 1qpj 1 jqpj 2 j 1q pi ` 1 i ` 2q. pi i`1qpi`1 i`2q pj 2 j 1qpj 1 jqpj 2 j 1q pi`1 i`2qpi i`1q IH pi i`1qpi`1 jqpi i`1q pi jq. Thus all transpositions are generated by adjacent transpositions, and so implying that S n xt 1 y. S n Ě xt 1 y Ě xt y S n, (b) Show that σ 2 is an even permutation for every σ P S n. Proof. Let ɛ : S n Ñ t 1uˆ be the sign function. Then for any σ P S n, since ɛ is a homomorphism, we have ɛpσ 2 q ɛpσq 2 p 1q 2 1. So σ 2 is even. (c) Let x, y be distinct 3-cycles in S n. (i) Set n 4 and assume x y 1. Show xx, yy A 4. Proof. Since x y 1 and xx, yy xx, y 1 y, we can assume without loss of generality that x and y are of the form where ta, b, c, du r4s. Then x pabcq and y pbcdq, 1, x, x 2 pacbq, y, y 2 pbdcq, yxy 1 pacdq, and yx 2 y 1 pcadq padcq are distinct elements of xx, yy, so that xx, yy ě 7. But since x and y are both even, we know xx, yy ď A 4, and so xx, yy divides A So xx, yy 12, and therefore xx, yy A 4. 1

2 2 (ii) Set n 5 and assume x y 1. Show that either x and y both fix some common elements of r5s (there is some i P r5s such that xpiq i and ypiq i) and xx, yy A 4, or x and y do not fix any common elements of r5s (for all i P r5s, if xpiq i then ypiq i) and xx, yy A 5. Proof. If x and y both fix some common element a P r5s, then if A r5s tau, we have x, y P S A S 4. So this case reduces to the setting of the previous part, and thus xx, yy A 4. Otherwise, using similar reasoning as in the previous case, without loss of generality, we can write x pabcq and y pcdeq, where ta, b, c, d, eu r5s. Again, 1 P xx, yy. And x, x 2, y, and x 2 are distinct 3-cycles contained in xx, yy. And we can generate all 20 3-cycles as follows, using our conjugation rule for permutations: pabcq x pacbq x 1 pcdeq y pcedq y 1 pabdq yxy 1 padbq pyxy 1q 1 pabeq y 1 xy paebq py 1xyq 1 padeq xyx 1 paedq pxyx 1q 1 pbdeq x 1 yx pbedq px 1yxq 1 paceq yxy 1x 1y 1 paecq pbcdq xyxy 1x 1 pbdcq pxyxy 1x 1q 1 pacdq x 1yx 1y 1 x padcq pbceq xy 1xyx 1 pbecq pxy 1xyx 1q 1 pyxy 1 x 1 y 1 q 1 px 1 yx 1 y 1 xq 1 Now using pairs x 1, y 1 of these three cycles that overlap at exactly two values (like x 1 pabcq and y 1 pbcdq) as in the previous part, we can also generate all ` permutations that are the product of two disjoint 2-cycles (since they are elements of the corresponding isomorphic image of A 4 ). So xx, yy ě t1u ` t3-cyclesu ` tproduct of two disjoint 2-cyclesu 1 ` 20 ` But since x and y are both even, we know xx, yy ď A 5, and so xx, yy divides A So xx, yy 60, and therefore xx, yy A 5. (iii) Show, for all n, that xx, yy is isomorphic to one of Z 3, A 4, A 5, or Z 3 ˆ Z 3. Proof. Writing x px 1 x 2 x 3 q, y py 1, y 2, y 3 q, and A tx 1, x 2, x 3, y 1, y 2, y 3 u, then we have xx, yy ď S A S A. Depending on how many terms x and y share, we have A 3, 4, 5, or 6. If A 3, then x y or x y 1, so that xx, yy xxy Z 3. If A 4 then x y, y 1, and xx, yy ď S A S 4, so that we are in the setting of part (a). So xx, yy A 4. Similarly, if A 5, then

3 we are in the second setting of part (b) and xx, yy A 5. If A 6, then x and y are disjoint, so that xy yx. So xx, yy Z 3 ˆ Z Group actions. (a) Let G ü A. Prove that if a, b P A and b g a for some g P G, then G b gg a g 1. Deduce that if G acts transitively on A, then the kernel of the action is Ş gpg gg ag 1. Proof. If b g a and x P G a, then pgxg 1 q b g px pg 1 g aqqq g px aq g a b so gg a g 1 Ď G b. Similarly, since b g a, we have a g 1 a, so that g 1 G b g Ď G a. This is true if and only if G b Ď gg a g 1. So G b gg a g 1. If G acts transitively on A, there is only one orbit, so that every b P A can be written as g b a for some g b P G. So ker č G b č g b G g 1 a b č gg g 1 a. bpa bpa gpg (b) Let S 3 act on the set of ordered triples A tpi, j, kq i, j, k P r3su. (i) Find the orbits of S 3 ü A. Answer. Note that for any σ P S n, i j if and only if σpiq σpjq (and similarly for i and k, and j and k). And for l P rns there is some σ P S n such that σpiq l (and similarly for j and k). So the orbits of the action on A are classified by which entries are the same and which aren t. For example, O p1,1,2q tpi, i, jq i, j P rns, i ju and O p1,2,1q tpi, j, iq i, j P rns, i ju. So there are five orbits in total, given by O p1,1,1q, O p1,1,2q, O p1,2,1q, O p2,1,1q, and O p1,2,3q. (ii) For each orbit O, choose one representative a P O and calculate G a. G : G a O. Verify that

4 4 Answer. a p1, 1, 1q : G a t1, p23qu, so G : G a 6{2 3 O a 3 a p1, 1, 2q : G a t1u, so G : G a 6{1 6 O a a p1, 2, 1q : G a t1u, so G : G a 6{1 6 O a a p2, 1, 1q : G a t1u, so G : G a 6{1 6 O a a p1, 2, 3q : G a t1u, so G : G a 6{1 6 O a (c) Briefly give an example of an action of a group G on a set A satisfying the following properties. (It s ok to give examples from class or the book.) (a) transitive and faithful Answer. G 1 acting on A tau trivially. (b) transitive but not faithful Answer. G Z 2 acting on A tau trivially. (c) faithful but not transitive Answer. G 1 acting on A ta, bu trivially. (d) not transitive or faithful Answer. G Z 2 acting on A ta, bu trivially. (e) transitive, faithful, and trivial Answer. G 1 acting on A tau trivially. (d) Suppose G acts transitively on a finite set A, and let H IJ G. Note that the action of G on A restricts to an action of H on A, which is not necessarily transitive anymore. Let O 1, O 2,..., O r be the distinct orbits of the action of H on A. (i) Show that for each a P A, H a G a X H.

5 5 Proof. We have H a th P H h a au tg P G g a au X H G a X H. (ii) Prove that G permutes O 1, O 2,..., O r, i.e. for each g P G, i P rrs, we hace g O i O j for some j P rrs (where g O i : tg a a P O i u); and σ g : to 1, O 2,..., O r u Ñ to 1, O 2,..., O r u defined by O i ÞÑ g O i is a bijection for each g P G. Proof. If we write O i H a i for some a i P O i, then since H is normal in G, we have ( ) g O i g ph a i q gh a i Hg a i H pg a i q. So since the orbits partition A and g a P A, we have that H pg aq is one of the orbits. And since left action by g 1 inverts the left action by g, σ g is a bijection. (iii) Deduce that G acts on the set A to 1, O 2,..., O r u. Show that this action is transitive, and deduce that O i O j for all i, j P rrs. Proof. Using ( ), we have 1 O i H p1 a i q H a i O i and g 1 pg 2 O i q g 1 H pg 2 a i q H pg 1 pg 2 a i qq H pg 1 g 2 a 1 q g 1 g 2 H a 1 g 1 g 2 O 1. So G acts on A. For all orbits H a, H b P A, since G acts transitively on A, there is some g P G such that b g a, so So the action on A is transitive. g ph aq H pg aq H b. Since the map from one orbit to the other given by O i ÞÑ g O i is a bijection (since it is invertible by left action of g 1 ) we have O i g O i for all g P G. But G acts transitively on A, so O i O j for all i, j P rrs. (iv) Fix O P A, and let a P O (so that O H a). Show that O H : H X G a and that r G : HG a (where r A as above). Proof. Since O H : H a and H a H X G a, it follows that O H : H X G a. Now let r A be the number of H-orbits. Since H IJ G, the second isomorphism theorem tells us that HG a ď G, and give us the following lattice, where the marked

6 6 edges (from subgroup A down to subgroup B) indicate the same index ( A : B ): H G HG a H X G a 1 G a So HG a : G a H : H X G a, and HG a : H G a : H X G a. And since, for any C ď B ď A, we have C : A A : B B : C. So G : G a G : HG a HG a : G a. Since the action of G is transitive, we have A G : G a. And since the H-orbits partition A, and every orbit has the same size, we have A r O for any O P A. So G : G a r O r H : H a r H : H X G a r HG a : G a. Therefore, since G : G a, r, and O are all finite, and G : G a G : HG a HG a : G a, we have r G : G a { HG a : G a G : HG a. 3. Left actions of groups on themselves (a) Label the elements of S 3 as label: x 1 x 2 x 3 x 4 x 5 x 6 element: 1 p12q p23q p13q p123q p132q Let S 3 act on itself by left multiplication (σ τ στ for σ, τ P S n ), and consider the maps ϕ : S 3 Ñ S S3 S tx1,...,x 6 u induced by the action of S 3 on itself, and ψ : S tx1,...,x 6 u Ñ S 6 defined by ψpσq : i ÞÑ j iff σ : x i ÞÑ x j. For example, since p12q x 1 p12q1 p12q x 2, p12q x 2 p12qp12q 1 x 1, p12q x 3 p12qp23q p123q x 5, p12q x 4 p12qp13q p132q x 6, p12q x 5 p12qp123q p23q x 3, p12q x 6 p12qp132q p13q x 4, we have ϕpp12qq is the map that sends x 1 ÞÑ x 2, x 2 ÞÑ x 1, x 3 ÞÑ x 5, x 4 ÞÑ x 6, x 5 ÞÑ x 3, x 6 ÞÑ x 4, and ψpϕpp12qqq is the map that sends 1 ÞÑ 2, 2 ÞÑ 1, 3 ÞÑ 5, 4 ÞÑ 6, 5 ÞÑ 3, 6 ÞÑ 4, so that ψpϕpp12qqq p12qp35qp46q. Give the images of the rest of the elements of S 3 under ψ ϕ, and verify that S 3 ψ ϕps 3 q (since ψ ϕ is injective). [It s ok to google multiplication table for S 3 to expedite this problem.]

7 Answer. We have the action of S 3 on tx 1,..., x 6 u is given by x 1 x 2 x 3 x 4 x 5 x 6 1 x 1 x 2 x 3 x 4 x 5 x 6 p12q x 2 x 1 x 5 x 6 x 3 x 4 p23q x 3 x 6 x 1 x 5 x 4 x 2 p13q x 4 x 5 x 6 x 1 x 2 x 3 p123q x 5 x 4 x 2 x 3 x 6 x 1 p132q x 6 x 3 x 4 x 2 x 1 x 5 so that ψ ϕ : 1 ÞÑ 1 p12q ÞÑ p12qp35qp46q p23q ÞÑ p13qp26qp45q p13q ÞÑ p14qp25qp36q p123q ÞÑ p156qp243q p132q ÞÑ p165qp234q Since ψ ϕ is a homomorphism and has 6 distinct elements in its image, we have S 3 ψ ϕps 3 q. (b) Consider H xsy inside of D 2 3, and label the cosets as label: x 1 x 2 x 3 element: H rh r 2 H. Repeat problem 1, listing the images of ψ ϕ : D 6 Ñ S 3 (where ϕ : D 6 Ñ S tgh gpd6 u S tx1,x 2,x 3 u and ψ : S tx1,x 2,x 3 u Ñ S 3 are defined analogously. Answer. We have H t1, su, rh tr, rsu tr, sr 2 u, r 2 H tr 2, r 2 su tr 2, sru. So the action of D 6 on tx 1, x 2, x 3 u is given by x 1 x 2 x 3 1 x 1 x 2 x 3 r x 2 x 3 x 1 r 2 x 3 x 1 x 2 s x 1 x 3 x 2 sr x 3 x 2 x 1 sr 2 x 2 x 1 x 3 so that ψ ϕ : 1 ÞÑ 1 r ÞÑ p123q r 2 ÞÑ p132q s ÞÑ p23q sr ÞÑ p13q sr 2 ÞÑ p12q Since ψ ϕ is a homomorphism and has 6 distinct elements in its image, we have D 6 ψ ϕpd 6 q S 3. (c) Show that if H has finite index n in G, then there is a normal subgroup K IJ G with K ď H and G : K ď n!. Proof. Let G act on the left cosets of H, of which there are n. This action corresponds to a homomorphism σ : G Ñ S A, where A tgh g P Gu. Let K kerpσq. We have So G : K σpgq ď S n n!. G{K σpgq ď S A S n. 7